L6 · Factoring x^2 + bx + c and difference of squares (integers)
“I can factor x^2 + bx + c by finding the integer pair that multiplies to c and adds to b, factor a difference of squares on sight, and recognize a trinomial that does not factor over the integers - verifying every answer by expanding.”
Factor x^2 + 7x + 12. Recall from L4 what the answer's expansion looks like: (x + p)(x + q) expands to x^2 + (p + q)x + pq. So the two constants we need must satisfy: pq = 12 (the constant) and p + q = 7 (the x-coefficient). Step 1: List factor pairs of 12: (1, 12) sum 13; (2, 6) sum 8; (3, 4) sum 7. Step 2: The pair (3, 4) hits both targets. Answer: (x + 3)(x + 4). Step 3 - verify by expanding: x^2 + 4x + 3x + 12 = x^2 + 7x + 12. Sign logic, stated once and used all lesson: if c is positive, p and q have the same sign, and b tells you which. If c is negative, p and q have opposite signs, and the one with the larger absolute value carries b's sign.
In x^2 + 7x + 12, why must the two constants multiply to exactly 12 and add to exactly 7?
Factor x^2 + 2x - 15. -> (x + 5)(x - 3) (c negative - opposite signs; pairs for 15: (1, 15), (3, 5); with sum +2, take +5 and -3. Check: x^2 - 3x + 5x - 15 = x^2 + 2x - 15)
"x^2 - 8x + 15 = (x - 3)(x - ___). What is the missing constant?" -> 5 (pair must multiply to +15 and add to -8: both negative, -3 and -5)
10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.